=( sqrt(3)^(log_(sqrt(3))16) )^(2*(x^2-1))=16^(2*(x^2-1))=
=(2^4)^(2*(x^2-1))=2^(8x^2-8)
32^(x-1)=(2^(5))^(x-1)=2^(5x-5)
(2^(8x^2-8) - 2^(5x-5))/(1-2x) ≤ 0
Умножим на 2^(8)
(2^(8x^2)-8*2^(5x))/(1-2x) ≤ 0
(2^(8x^2)-8*2^(5x))/(2x -1 ) ≥ 0
Применяем обобщенный метод интервалов
2^(8x^2)-8*2^(5x)=0
(2^x)^(8x) -8*(2^(x))^5=0
2^(x)=t
t^(8x)-8t^5=0
t^(8x)=8t^5
8x=log_(t)8t^5
8x=log_(t)8+5
8x=log_(2^(x))8+5
8x=(3/x)+5
x=1 или x=-3/8
____ [-3/8] __+__ (1/2) ____ [1] __+__
[-3/8;1/2)U[1;+ ∞)